The bottom of a cylindrical vessel has a hole of area $A$. If water is filled up to a height $h$,it flows out in $t$ seconds. If water is filled to a height $4h$,it will flow out in time:

In a cylindrical vessel containing liquid of density $\rho$,there are two holes in the side walls at heights of $h_1$ and $h_2$ respectively,such that the range of efflux at the bottom of the vessel is the same. The height of a hole,for which the range of efflux would be maximum,is:

$A$ tank is filled with water of density $1 \, g/cm^3$ and oil of density $0.9 \, g/cm^3$. The height of the water layer is $100 \, cm$ and the height of the oil layer is $400 \, cm$. If $g = 980 \, cm/s^2$,then the velocity of efflux from an opening in the bottom of the tank is

$A$ box filled with water has a small hole on its side near the bottom. It is dropped from the top of a tower. As it falls,a camera attached on the side of the box records the shape of the water stream coming out of the hole. The resulting video will show

$A$ large tank open to the atmosphere at the top and filled with water develops a small hole in the side at a point $20 \,m$ below the water level. If the rate of flow of water from the hole is $3 \times 10^{-3} \,m^3/min$, then the area of the hole is (Acceleration due to gravity $= 10 \,m/s^2$) (in $\,mm^2$)

$A$ vertical tank,open at the top,is filled with a liquid and rests on a smooth horizontal surface. $A$ small hole is opened at the centre of one side of the tank. The area of cross-section of the tank is $N$ times the area of the hole,where $N$ is a large number. Neglect the mass of the tank itself. The initial acceleration of the tank is: